
Vaccine hesitancy threatens to reverse the progress in tackling vaccine-preventable diseases. We used an SIS model with a game theory model for vaccination and parameters from the COVID-19 pandemic to study how vaccine hesitancy impacts epidemic dynamics. The system showed three asymptotic behaviors: total rejection of vaccinations, complete acceptance, and oscillations. With increasing fear of infection, stable endemic states become periodic oscillations. Our results suggest that managing fear of infection relative to vaccination is vital to successful mass vaccinations.
Citation: Anthony Morciglio, R. K. P. Zia, James M. Hyman, Yi Jiang. Understanding the oscillations of an epidemic due to vaccine hesitancy[J]. Mathematical Biosciences and Engineering, 2024, 21(8): 6829-6846. doi: 10.3934/mbe.2024299
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Vaccine hesitancy threatens to reverse the progress in tackling vaccine-preventable diseases. We used an SIS model with a game theory model for vaccination and parameters from the COVID-19 pandemic to study how vaccine hesitancy impacts epidemic dynamics. The system showed three asymptotic behaviors: total rejection of vaccinations, complete acceptance, and oscillations. With increasing fear of infection, stable endemic states become periodic oscillations. Our results suggest that managing fear of infection relative to vaccination is vital to successful mass vaccinations.
Rotating blades (thin-walled beam) are important structures widely used in mechanical and aerospace engineering as aviation engine blades, various cooling fans, windmill blades, helicopter rotor blades, airplane propellers etc. The study of the dynamics of rotating blades is important to design purposes, optimization, and control.
If the shear effect is not considered, the Euler-Bernoulli beam equation is used to model vibration of thin-walled beam. Chen et al.[6] studied the boundary feedback stabilization of a linear Euler-Bernoulli beam equation, they proved that the total energy of the equation decays uniformly and exponentially with
A very extensive work devoted to the longitudinal vibration in two directions were done by Librescu and Song[21,32,33] and their co-workers[25,28]. Under assumption of the cross-section to be rigid in its own plane, they modelled the rotating blades by 1-D linear governing equations. The influence of many factors on rotating blades, such as the anisotropy and heterogeneity of constituent materials, functionally graded materials (FGM), temperature, shear effects, primary and secondary warping phenomena (Vlasov effect), centrifugal and Coriolis forces etc have been taken into account in the 1-D linear governing equations.
Following Librescu's approach, various blades models were derived. Georgiades et al.[11] modelled a rotating blades by means of linear strain-displacement relationships, considering arbitrary pitch (presetting) angle and non-constant rotating speed. Choi et al.[8] studied bending vibration control of the pre-twisted rotating composite blades, who emphasized the important of piezoelectric effect in single cell composite blades. Fazelzadeh et al.[9] considered a thin-walled blades made of FGM which is used in turbomachinery under aerothermodynamics loading. In the paper, quasi-steady aerodynamic pressure loadings was determined by the first-order piston theory, and steady beam surface temperature was obtained from gas dynamics theory. Fazelzadeh et al.[10] studied the governing equations which included the effects of the presetting angle and the rotary inertia. The effects of steady wall temperature and quasi-steady aerodynamic pressure loadings due to flow motion were also taken into account.
The models in Refs.[21,32,33,25,28,11,8,9,10] are linear. When the engine blades rotate at a low speed, the linear approximation can completely meet the needs of practical application. However, when the blade rotate at a high speed, the simple linear approximation can not accurately describe the dynamic behavior of the system. So the non-linear analysis of rotating blades has attracted considerable attention.
The nonlinear governing equations of a rotating blades at constant angular velocity was presented by Anderson[1], and the author linearized the equation under the assumption that a small perturbed motion occurred at an initially stressed equilibrium configuration. Chen et al.[7] considered the effects of geometric non-linearity, shear deformation and rotary inertia. Arvin et al.[2] builded a nonlinear governing equation for rotating blades considering centrifugal forces by means of von-Karmans strain-displacement relationships under assumption of the constant rotation speed and zero pitch (presetting) angle. Yao et al.[39] employed the Hamilton's principle to derive the nonlinear governing equations with periodic rotating speed, arbitrary pitch (presetting) angle and linear pre-twist angle. Under the assumption that the location of shear centre is different from the centre of gravity, Avramov et al.[3] obtained results of the investigations on flexural-flexural-torsional nonlinear vibrations of twisted rotating blades described by the model of three nonlinear integro-differential equations. Other nonlinear models can be found in Refs.[34,29,27,37,15,31,36].
To the best of the author's knowledge, all the above literatures about the longitudinal vibration in two directions skipped the existence proof of solutions to the governing equations, and directly used the finite element method to study the influence of various parameters on blades vibration. To address this situation, we first try to model a governing equations of the blades with arbitrary rotating speed, arbitrary pre-setting angle and pre-twist angle. In the process of building the model, we take into account the free vibration at the right end of the blade and the nonlinear relationship between stress and strain. In the paper we aim to investigate the well-posedness and regularity of the governing equations. The well-posedness of other nonlinear blade vibration models can be found in Refs. [5,38,26,35,16].
Let us consider a slender, straight blades mounted on a rigid hub of radius
To derive the model of the rotating blades, the following kinematic and static assumptions are postulated:
(ⅰ) The blades is perfectly elastic bodies, the blades material is isotropic and is not affected by temperature,
(ⅱ) The cross section of the blades is rectangular and all its geometrical dimensions remain invariant in its plane,
(ⅲ) The ratio of wall thickness
(ⅳ) The transverse shear effect of the cross section is neglected,
(ⅴ) The axial displacement
Inertial Cartesian coordinate systems
Rotating coordinate systems
Transformation between
(X,Y,Z)=(x,y,z)+(R0,0,0). | (1) |
Local coordinates systems
(1000cosω−sinω0sinωcosω). |
Local, curvilinear coordinate systems
In order to determine the relationship between the two coordinate systems
r=xi+y(s)j+z(s)k. | (2) |
The position vector
r∗=r+nen, | (3) |
As a result
et=drds=dy(s)dsj+dz(s)dsk, | (4) |
en=et×i=dz(s)dsj−dy(s)dsk. | (5) |
Moreover, in order to avoid confusion, the notation
x∗=x, y∗=r∗⋅j=y+ndzds, z∗=r∗⋅k=z−ndyds. | (6) |
Based on the assumptions (ⅳ) and (ⅴ), the axial displacement
Dx=ϕy(x,t)(z(s)−ndyds)+ϕz(x,t)(y(s)+ndzds),Dy=u,Dz=v. | (7) |
where
ϕy=−vx,ϕz=−ux. | (8) |
where
Based on the assumptions (ⅴ), the displacement-strain relationships is expressed as follows[20]:
εxx=∂Dx∂x+12[(∂Dy∂x)2+(∂Dz∂x)2]. | (9) |
Thanks to (8), we can get
εxx=ˉεxx+ˉˉεxxn, |
where
ˉεxx=12(u2x+v2x)−(uxxy(s)+vxxz(s)), ˉˉεxx=−uxxdzds+vxxdyds. | (10) |
The shear strain
εnx=12(∂Un∂x+∂w∂n), | (11) |
where
Un=(Dx,Dy,Dz)⋅en=udzds−vdyds. | (12) |
Substituting (7) and (12) into (11), we deduce
εsx=12(dydsγxy+dzdsγxz)=0. | (13) |
where
Since these materials are isotropic, the corresponding thermoelastic constitutive law adapted to the case of structures is expressed as
(σssσxxσxnσnsσsx)=(Q11Q12000Q21Q2200000Q4400000Q5500000Q66)(εssεxxεxnεnsεsx), |
Herein, the reduced thermoelastic coefficients are defined as:
Q11=Q22=E1−ν2,Q12=Q21=Eν1−ν2,Q44=Q55=k2E2(1+ν),Q66=E2(1+ν), |
where
According to assumption (ⅱ), the cross section is rigid, then we can derive
(σss,σxx,σxn,σns,σsx)=(0,Eεxx,0,0,0). |
The centrifugal force can be represented as
Fc=∫lxρˉAω2(R0+x)dx=ρˉAω2R(x), |
where
In the paper, we use the first-order piston theory (see Ref.[24]) to evaluate the perturbed gas pressure. The pressure on the principal plane of the blade can be obtained as
Pyp=C∞ρ∞(∂up∂t+Utyp∂up∂x),Pzp=C∞ρ∞(∂vp∂t+Utzp∂vp∂x), | (14) |
where
Utyp=U∞cosθ, Utzp=U∞sinθ, | (15) |
The transformation relationship between
up=ucosθ+vsinθ, vp=−usinθ+vcosθ. | (16) |
Therefore, the external forces per unit axial length in the
py=aPzpsinθ−bPypcosθ, pz=−aPzpcosθ−bPypsinθ. | (17) |
Combining (14), (15), (16),
py=b1ux+b2vx+b3u+b4v+b5ut+b6vt | (18) |
pz=e1ux+e2vx+e3u+e4v+e5ut+e6vt | (19) |
where
b1=−C∞ρ∞U∞(asin3θ+bcos3θ)b2=−C∞ρ∞U∞sinθcosθ(−asinθ+bcosθ)b3=−C∞ρ∞U∞θxsinθcosθ(asinθ−bcosθ)b4=−C∞ρ∞U∞θx(asin3θ+bcos3θ)b5=−C∞ρ∞(asin2θ+bcos2θ)b6=C∞ρ∞sinθcosθ(a−b)e1=−C∞ρ∞U∞sinθcosθ(−asinθ+bcosθ)e2=−C∞ρ∞U∞sinθcosθ(acosθ+bsinθ)e3=C∞ρ∞U∞θxsinθcosθ(asinθ+bcosθ)e4=−C∞ρ∞U∞θxsinθcosθ(−asinθ+bcosθ) |
e5=C∞ρ∞sinθcosθ(a−b)e6=−C∞ρ∞(acos2θ+bsin2θ) |
In order to calculate the kinetic energy, the velocity vector and the acceleration should be given first. Based on the assumption (ⅴ), the position vector
R(X,Y,Z,t)=x∗i+(y∗+u)j+(z∗+v)k+R0i. |
Keep in mind that the rotation takes place solely in the
it=ωj; jt=−ωi; kt=0. |
Then the velocity vector and the acceleration of an arbitrary point
Rt=−ω(y∗+u)i+(ω(R0+x)+ut)j+vtk, | (20) |
Rtt=−[2ωut+ωt(y∗+u)+ω2(R0+x)]i +[utt−ωt(R0+x)−ω2(y∗+u)]j+vttk, | (21) |
where subscript
In order to derive the blades model and the associated boundary conditions, the extended Hamilton's principle is used. This can be formulated as
∫t2t1(δK−δU+δW)dt=0,δu=0,δv=0att=t1,t2. | (22) |
where
Thanks to the cross section of the blades is rectangular, we get
∮y(s)ds=∮z(s)ds=0. | (23) |
Utilizing (23), the kinetic energy is obtained
K=12∫τρR2tdτ=12ρ∫τu2t+v2t+2ω(R0+x)ut +ω2(u2+2(y+ndzds)u+y+ndzds+(R0+x)2)dτ=12ρˉA∫l0u2t+v2t+2ω(R0+x)ut +ω2(u2+2h∮ydsˉAu+h∮ydsˉA+(R0+x)2)dx=12ρˉA∫l0u2t+v2t+2ω(R0+x)ut+ω2(u2+(R0+x)2)dx, |
where
∫t2t1δKdt=−ρˉA∫t2t1∫l0{[utt+ωt(R0+x)−uω2]δu+vttδv}dxdt, | (24) |
Due to the rotating motion of the blades, the total strain energy consists of two parts. The strain energy caused by the centrifugal force can be obtained as
U1=12∫l0∮∫h2−h2∫lxρω2(R0+ς)εxxdςdndsdx | (25) |
where
δU1=12∫l0∮∫h2−h2∫lxρω2(R0+ς)δεxxdςdndsdx=12∫l0∮∫h2−h2∫lxρω2(R0+ς)δˉεxxdςdndsdx=ρˉAω2∫l0R(x)(uxδux+vxδvx)dx−ρω2h2∫l0∫lx(R0+ς)(∮y(s)dsδuxx+∮z(s)dsδvxx)dςdx, | (26) |
where
Thanks to (23), the variation of the strain energy caused by the centrifugal force can be rewritten as
δU1=ρˉAω2[R(x)(uxδu+vxδv)]|l0−∫l0(R(x)ux)xδu+(R(x)vx)xδvdx | (27) |
The strain energy induced by the deformation of the rotating blades can be expressed as
U2=12∫l0∮∫h2−h2σxxεxxdndsdx. | (28) |
Substituting the expressions of the stress and strain resultants into (28) yields
δU2=E∫l0∮∫h2−h2εxxδεxxdndsdx=Eh4∫l0∮(u2x+v2x)δ(u2x+v2x)dsdx−Eh2∫l0∮(u2x+v2x)δ(uxxy(s)+vxxz(s))dsdx−Eh2∫l0∮(uxxy(s)+vxxz(s))δ(u2x+v2x)dsdx+Eh∫l0∮(uxxy(s)+vxxz(s))δ(uxxy(s)+vxxz(s))dsdx+Eh312∫l0∮(−uxxdzds+vxxdyds)δ(−uxxdzds+vxxdyds)dsdx | (29) |
δU2=E∫l0∮∫h2−h2εxxδεxxdndsdx=ρˉA{∫l0[−a52((u2x+v2x)ux)x+(a6uxx−a3vxx)xx −a12(u2x+v2x)xx+(a1uxx+a2vxx)ux)x]dx}δu+ρˉA{∫l0[−a52((u2x+v2x)vx)x+(a4vxx−a3uxx)xx −a22(u2x+v2x)xx+(a1uxx+a2vxx)vx)x]dx}δv+ρˉA{[(a6uxx−a3vxx)−a12(u2x+v2x)]δux}|l0+ρˉA{[(a4vxx−a3uxx)−a22(u2x+v2x)]δvx}|l0+ρˉA{[a52(u2x+v2x)ux−(a6uxx−a3vxx)x +a12(u2x+v2x)x+a1uxx+a2vxxux]δu}|l0+ρˉA{[a52(u2x+v2x)vx−(a4vxx−a3uxx)x +a22(u2x+v2x)x+a1uxx+a2vxxvx]δv}|l0 | (30) |
where
a1=EhˉAρ∮yds=0;a2=EhˉAρ∮zds=0;a3(x)=EhˉAρ∮h212dydsdzds−yzds;a4(x)=EhˉAρ∮h212(dzds)2+y2ds;a5=Eρa6(x)=EhˉAρ∮h212(dyds)2+z2ds. |
The work of the non-conservative external forces can be obtained as
W=∫l0pyu+pzvdx. | (31) |
Then
∫t2t1δWdt=∫t2t1∫l0˜pyδu+˜pzδvdxdt+∫t2t1{(b1u+e1v)δu+(b2u+e2v)δv}|l0dt. | (32) |
where
˜py=2b3u+(e3+b4)v | (33) |
˜pz=(e3+b4)u+2e4v | (34) |
Inserting variation of potential energy (27) and (30), variation of kinetic energy (24) and variation of external work equation (32) into the extended Hamilton's principle (22), collecting the terms associated with the same variations, invoking the stationarity of the functional within the time interval
utt+(a6uxx−a3vxx)xx−a52((u2x+v2x)ux)x −ω2(R(x)ux)x−ω2u−¯p1+ωt(R0+x)=0, | (35) |
vtt+(a4vxx−a3uxx)xx−a52((u2x+v2x)vx)x−ω2(R(x)vx)x−¯p2=0, | (36) |
where
¯p1=˜pyρˉA, ¯p2=˜pzρˉA. | (37) |
The following two types of boundary conditions are generated due to the different design of engine blades.
u,ux=0,v,vx=0,x=0, | (38) |
u,ux=0,v,vx=0,x=l. | (39) |
and
u,ux=0,v,vx=0,x=0, | (40) |
a6uxx−a3vxx=0,x=l, | (41) |
a4vxx−a3uxx=0,x=l, | (42) |
a52(u2x+v2x)ux−(a6uxx−a3vxx)x−b1u−e1v=0,x=l, | (43) |
a52(u2x+v2x)vx−(a4vxx−a3uxx)x−b2u−e2v=0,x=l. | (44) |
The conditions (38)-(39) represent C-C boundary condition. The conditions (40)-(44) represent C-F boundary condition. Now we study the well-posedness and regularity of the solution for C-C and C-F blades.
We write
H20={ψ∈H2(Ω)|ψ(x)=0,ψx(x)=0,x∈∂Ω},H2f={ψ∈H2(Ω)|ψ(0)=0,ψx(0)=0,x=0}. |
We list Gagliardo-Nirenberg inequality for bounded domains (see [23]) to be used in the subsequent sections
Lemma 3.1. Let
1p−jn=α(1r−mn)+(1−α)1q,jm≤α≤1 |
Then
‖Djf‖Lp(Ω)≤C1‖Dmf‖αLr(Ω)‖f‖1−αLq(Ω)+C2‖f‖Ls(Ω) |
where
Now, we give the Aubin-Lions Lemma (see [22]).
Lemma 3.2. Suppose
(i)
(ii)
(iii)
then
Without loss of generality, we assume
{utt+(a6uxx)xx−(a3vxx)xx−12((u2x+v2x)ux)x −ω2(Rux)x−p1=0inQ,vtt−(a3uxx)xx+(a4vxx)xx−12((u2x+v2x)vx)x −ω2(Rvx)x−p2=0inQ,u,ux=0,v,vx=0on∂Ω×[0,T],u=u0(x),ut=u1(x),v=v0(x),vt=v1(x)onΩ×{t=0}. | (45) |
where
p1=¯p1+ω2u−ωt(R0+x), p2=¯p2. | (46) |
Definition 4.1. We say function
ut,vt∈L∞(0,T;L2(Ω)),utt,vtt∈L∞(0,T;H−2(Ω)), |
is a weak solution of the initial boundary value problem (45) provided
(ⅰ)
(utt,φ)+(a6uxx,φxx)−(a3vxx,φxx)+12((u2x+v2x)ux,φx)+ω2(Rux,φx)−(p1,φ)=0. | (47) |
(vtt,φ)−(a3uxx,φxx)+(a4vxx,φxx)+12((u2x+v2x)vx,φx)+ω2(Rvx,φx)−(p2,φ)=0. | (48) |
for each
(ⅱ)
u(0)=u0,ut(0)=u1;v(0)=v0,vt(0)=v1. | (49) |
Remark 1. From the definition, we know
Remark 2. By the product Minkowshi inequality, we can obtain:
a4a6−a23>0,x∈Ω. | (50) |
Theorem 4.2. (Existence for weak solution of (45)) Assume
ω∈C1(0,T),a3,a4,a6∈L∞(Ω),u0,v0∈H20(Ω),u1,v1∈L2(Ω). | (51) |
there exists a weak solution of (45).
We now briefly outline the proof of Theorem 4.2 in the following:
Step 1. employing Galerkin's method to construct solutions of certain finite-dimensional approximations to (45) (correspond to Lemma 6.1 in chapter 6);
Step 2. using the energy method to find the uniform estimates of the finite-dimensional approximations solutions (correspond to Lemma 6.2 in chapter 6);
Step 3. using compactness method to obtain the weak solutions of (45).
Now we give the smoothness of weak solutions of (45).
Theorem 4.3. (Improved regularity) Assume
{a3,a4,a6∈L∞(Ω),ω∈C1(0,T),ωtt∈L∞(0,T),u0∈H20(Ω)∩H4(Ω),v0∈H20(Ω)∩H4(Ω),u1∈H20(Ω),v1∈H20(Ω), | (52) |
the weak solution of (45) satisfies
ut,vt∈L∞(0,T;H20(Ω)),utt,vtt∈L∞(0,T;L2(Ω)). | (53) |
Theorem 4.4. (Interior regularity) Under the condition (52), for any
φuxxx,φvxxx∈L∞(0,T;L2(Ω)). |
Remark 3. Multiplying the first and second equation of (45) by
uxxxx,vxxxx∈L∞(0,T;L2(Ω′)). |
where
If the pre-twist angle
Theorem 4.5. (Improved regularity when
u,v∈L∞(0,T;H4(Ω)∪H20(Ω)). |
Now, we study the uniqueness and stability of (45), denote operator
π:{a3,a4,a6,ω,θ,u0,v0,u1,v1}→{u,v}, |
and
W(Q)={ψ|ψ∈L∞(0,T;H20(Ω)),ψt∈L∞(0,T;L2(Ω))} |
with the norm
||ψ||W(Q)=||ψ||L∞(0,T;H20(Ω))+||ψt||L∞(0,T;L2(Ω)). |
Then
Theorem 4.6. Undering the condition (52),
π:{(L∞(Ω))3×W1,1(0,T)∩C1(0,T)×L1(Ω)×(H20(Ω))2×(L2(Ω))2}→(W(Q))2 |
is continuous.
The initial boundary-value problems of C-F blades are rewritten as :
{utt+(a6uxx)xx−(a3vxx)xx−12((u2x+v2x)ux)x −ω2(Rux)x−p1=0inQ,vtt−(a3uxx)xx+(a4vxx)xx−12((u2x+v2x)vx)x −ω2(Rvx)x−p2=0inQ,u,ux=0,v,vx=0,x=0,a6uxx−a3vxx=0,x=l,a4vxx−a3uxx=0,x=l,12(u2x+v2x)ux−(a6uxx−a3vxx)x−b1u−e1v=0,x=l,12(u2x+v2x)vx−(a4vxx−a3uxx)x−b2u−e2v=0,x=l,u=u0(x),ut=u1(x),v=v0(x),vt=v1(x)onΩ×{t=0}. | (54) |
Definition 5.1. We say function
ut,vt∈L∞(0,T;L2(Ω)),utt,vtt∈L∞(0,T;H−2(Ω)), |
is a weak solution of the initial boundary value problem (54) provided
(ⅰ)
(utt,φ)+(a6uxx,φxx)−(a3vxx,φxx)+12((u2x+v2x)ux,φx)+ω2(Rux,φx)−(p1,φ)−(b1u(l)+e1v(l))φ(l)=0. | (55) |
(vtt,φ)−(a3uxx,φxx)+(a4vxx,φxx)+12((u2x+v2x)vx,φx)+ω2(Rvx,φx)−(p2,φ)−(b2u(l)+e2v(l))φ(l)=0. | (56) |
for each
(ⅱ)
u(0)=u0,ut(0)=u1;v(0)=v0,vt(0)=v1. |
Similar as above, we derive the following conclusion of the initial boundary value problem (54).
Theorem 5.2. (Existence for the weak solutions of (54)) Assume
ω∈C1(0,T),a3,a4,a6∈L∞(Ω),u0,v0∈H2f(Ω),u1,v1∈L2(Ω), | (57) |
there exists a weak solution of (54).
Theorem 5.3. (Regularity weak solution of (54)) Assume
{a3,a4,a6∈L∞(Ω),ω∈C1(0,T),ωtt∈L∞(0,T),u0,v0∈H2f(Ω)∩H4(Ω),u1,v1∈H2f(Ω). | (58) |
Then there exists
ut,vt∈L∞(0,T∗;H2f(Ω)),utt,vtt∈L∞(0,T∗;L2(Ω)). | (59) |
Remark 4. If pre-twist angle
u,v∈L∞(0,T∗;H4(Ω)∪H2f(Ω)). | (60) |
where
We construct weak solution of the initial boundary-value problem (45) by first solving a finite dimensional approximation. We thus employ Galerkin's method by selecting smooth functions
{φk}∞k=1is an orthogonal basis ofH20(Ω), | (61) |
and
{φk}∞k=1is an orthonormal basis ofL2(Ω). | (62) |
Fix a positive integer
um=m∑k=1dk0m(t)φk, vm=m∑k=1dk1m(t)φk, | (63) |
where we intend to select the coefficients
dk0m(0)=(u0,φk),dk1m(0)=(v0,φk),k=1,⋯,m, | (64) |
dk0m,t(0)=(u1,φk),dk1m,t(0)=(v1,φk),k=1,⋯,m, | (65) |
and
(um,tt,φk)+(a6um,xx,φk,xx)−(a3vm,xx,φk,xx)+12((u2m,x+v2m,x)um,x,φk,x)+ω2(Rum,x,φk,x)−(p1m,φk)=0, | (66) |
(vm,tt,φk)−(a3um,xx,φk,xx)+(a4vm,xx,φk,xx)+12((u2m,x+v2m,x)vm,x,φk,x)+ω2(Rvm,x,φk,x)−(p2m,φk)=0. | (67) |
where
p1m=1ρˉA(2b3um+(e3+b4)vm)+ω2um−ωt(R0+x),p2m=1ρˉA((e3+b4)um+2e4vm). |
In order to proof Theorem 4.2, we need the following Lemma.
Lemma 6.1. Under the condition (51), for each integer
Proof. Assuming
dk0m,tt+m∑j=1dj0m(a6φj,xx,φk,xx)−m∑j=1dj1m(a3φj,xx,φk,xx)+12m∑j=1dj0m(((m∑i=1di0mφi,x)2+(m∑i=1di1mφi,x)2)φj,x,φk,x)+ω2m∑j=1dj0m(Rφj,x,φk,x)−1ρˉA(2b3dk0m+(e3+e4)dk1m)=ω2dk0m−ωt((R0+x),φk), | (68) |
dk1m,tt+m∑j=1dj1m(a4φj,xx,φk,xx)−m∑j=1dj0m(a3φj,xx,φk,xx)+12m∑j=1dj1m(((m∑i=1di0mφi,x)2+(m∑i=1di1mφi,x)2)φj,x,φk,x)+ω2m∑j=1dj1m(Rφj,x,φk,x)−1ρˉA((e3+e4)dk0m+2e4dk1m)=0. | (69) |
subject to the initial conditions (64), (65). According to standard theory for ODE, there exists unique
We propose now to send
Lemma 6.2. Undering the condition (51), there exists positive constant
||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H20(Ω)+||vm||2H20(Ω)≤C(||u1||2L2(Ω)+||v1||2L2(Ω)+||u0||2H20(Ω)+||v0||2H20(Ω))+C | (70) |
for
Proof. Multiplying equality (66) by
(um,tt,um,t)+(a6um,xx,um,txx)−(a3vm,xx,um,txx)+12((u2m,x+v2m,x)um,x,um,tx)+ω2(Rum,x,um,tx)−(p1m,um,t)=0 | (71) |
for a.e.
Multiplying equality (67) by
(vm,tt,vm,t)−(a3um,xx,vm,txx)+(a4vm,xx,vm,txx)+12((u2m,x+v2m,x)vm,x,vm,tx)+ω2(Rvm,x,vm,tx)−(p2m,vm,t)=0 | (72) |
for a.e.
To simplify the equation (71) and (72), we can get
12ddt‖um,t‖2L2(Ω)+12ddt‖√a6um,xx‖2L2(Ω)−(a3vm,xx,um,txx)+12((u2m,x+v2m,x)um,x,um,tx)+ω22ddt‖√Rum,x‖2L2(Ω)−(p1m,um,t)=0. | (73) |
12ddt‖vm,t‖2L2(Ω)+12ddt‖√a4vm,xx‖2L2(Ω)−(a3um,xx,vm,txx)+12((u2m,x+v2m,x)vm,x,vm,tx)+ω22ddt‖√Rvm,x‖2L2(Ω)−(p2m,vm,t)=0. | (74) |
Summing the equations (73) and (74), we discover
12ddt{||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||√a6um,xx||2L2(Ω)+||√a4vm,xx||2L2(Ω)+ω22(||√Rum,x||2L2(Ω)+||√Rvm,x||2L2(Ω))}−ddt||a3um,xxvm,xx||L1(Ω)+18ddt||u2m,x+v2m,x||2L2(Ω)=ωωt(||√Rum,x||2L2(Ω)+||√Rvm,x||2L2(Ω))+(p1m,um,t)+(p2m,vm,t). | (75) |
Since
12ddt{||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||√a6um,xx||2L2(Ω)+||√a4vm,xx||2L2(Ω)+ω22(||√Rum,x||2L2(Ω)+||√Rvm,x||2L2(Ω))}−ddt||a3um,xxvm,xx||L1(Ω)+18ddt||u2m,x+v2m,x||2L2(Ω)≤C(||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H10(Ω)+||vm||2H10(Ω))+C. | (76) |
where we used Young inequality.
Integrating (76) with respect to
||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||√a6um,xx||2L2(Ω)+||√a4vm,xx||2L2(Ω)−2||a3um,xxvm,xx||L1(Ω)+ω22(||√Rum,x||2L2(Ω)+||√Rvm,x||2L2(Ω))+14||u2m,x+v2m,x||2L2(Ω)≤C(Ω,T){||u1m||2L2(Ω)+||v1m||2L2(Ω)+||u0m||2H2(Ω)+||v0m||2H2(Ω)}+C∫t0||um,t||2L2(Ω)+||vm,t||2L2(Ω)dt+||um||2H10(Ω)+||vm||2H10(Ω)+C. | (77) |
Thanks to (50), there exists a constant
||√a6um,xx||2L2(Ω)+||√a4vm,xx||2L2(Ω)−2||a3um,xxvm,xx||L(Ω)≥C(||um||2H20(Ω)+||vm||2H20(Ω)). | (78) |
Substituting (78) into the inequality (77), By using Poincaré inequality, we find
||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H20(Ω)+||vm||2H20(Ω) ≤C(||u1m||2L2(Ω)+||v1m||2L2(Ω)+||u0m||2H20(Ω)+||v0m||2H20(Ω))+C∫t0||um,t||2L2(Ω)+||vm,t||2L2(Ω)dt+||um||2H20(Ω)+||vm||2H20(Ω)+C. | (79) |
Then, by using Gronwall inequality, we obtain
||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H20(Ω)+||vm||2H20(Ω)≤C(||um,t(0)||2L2(Ω)+||vm,t(0)||2L2(Ω)+||um(0)||2H20(Ω)+||vm(0)||2H20(Ω))+C≤C(||u1||2L2(Ω)+||v1||2L2(Ω)+||u0||2H20(Ω)+||v0||2H20(Ω))+C |
for
Remark 5. From the Energy estimates, we can obtain:
Thanks to Lemma 6.1 and Lemma 6.2, we can obtain the existence for the weak solutions of the initial boundary value problem (45).
Proof. (ⅰ) According to the energy estimates (70), we see that
{um}∞m=1,{vm}∞m=1isboundedinL∞(0,T;H20(Ω)); | (80) |
{um,t}∞m=1,{vm,t}∞m=1isboundedinL∞(0,T;L2(Ω)); | (81) |
{um,tt}∞m=1,{vm,tt}∞m=1isboundedinL∞(0,T;H−2(Ω)). | (82) |
As a consequence there exists subsequence
{uμ→u,vμ→vweakly∗in L∞(0,T;H20(Ω))uμ,t→ut,vμ,t→vtweakly∗in L∞(0,T;L2(Ω))uμ,tt→utt,vμ,tt→vttweakly∗in L∞(0,T;H−2(Ω)). | (83) |
(ⅱ) By Gagliardo-Nirenberg inequality, we can see
||uμ,x||L∞(Q)≤C||uμ,xx||L∞(0,T;L2(Ω))≤C. | (84) |
Otherwise,
||u2μ,x+v2μ,x||L∞(0,T;L2(Ω))≤||uμ,x||2L∞(0,T;L4(Ω))+||vμ,x||2L∞(0,T;L4(Ω))≤C||uμ||2L∞(0,T;H20(Ω))+c||vμ||2L∞(0,T;H20(Ω))≤C. | (85) |
Combining (84) and (85), we discover
||(u2μ,x+v2μ,x)uμ,x||L∞(0,T;L2(Ω))≤C||u2μ,x+v2μ,x||L∞(0,T;L2(Ω))≤C. | (86) |
Moreover, there exists a function
(u2μ,x+v2μ,x)uμ,x→χ weakly * in L∞(0,T;L2(Ω)). | (87) |
By Lemma 3.2, we can find
uμ→u,vμ→v strongly in L2(0,T;H10(Ω)). | (88) |
And so
uμ,x→ux,vμ,x→vx strongly in L2(Q), | (89) |
Thus
(u2μ,x+v2μ,x)uμ,x→(u2x+v2x)ux. | (90) |
Combining (87) and (90), we can obtain
(u2μ,x+v2μ,x)uμ,x→(u2x+v2x)ux weakly in L2(Q), | (91) |
where we used the Lemma 1.3 of Chapter 1 in [22]. Furthermore, we have
Meanwhile,
((u2μ,x+v2μ,x)uμ,x,φk,x)→((u2x+v2x)ux,φk,x) weakly * in L∞(0,T). | (92) |
In the same way,
((u2μ,x+v2μ,x)vμ,x,φk,x)→((u2x+v2x)vx,φk,x) weakly * in L∞(0,T). | (93) |
Next fix an integer
(uμ,tt,φk)+(a6uμ,xx,φk,xx)−(a3vμ,xx,φk,xx)+12((u2μ,x+v2μ,x)uμ,x,φk,x)−ω2(Ruμ,x,φk,x)−(p1μ,φk)=0. | (94) |
Thanks to (83), we can get
{(uμ,tt,φk)→(utt,φk)weakly∗inL∞(0,T),(a6uμ,xx,φk,xx)→(a6uxx,φk,xx)weakly∗inL∞(0,T),(a3vμ,xx,φk,xx)→(a3vxx,φk,xx)weakly∗inL∞(0,T),(p1μ,φk)→(p1,φk)weakly∗inL∞(0,T). | (95) |
From (92) and (95), we can discover
+v2x)ux,φk,x)+ω2(Rux,φk,x)−(p1,φk)=0 | (96) |
for all fixed
Note
{φk}∞k=1 isanorthogonalbasisofH20(Ω), |
then,
(utt,φ)+(a6uxx,φxx)−(a3vxx,φxx)+12((u2x+v2x)ux,φx)+ω2(Rux,φx)−(p1,φ)=0 | (97) |
for arbitrary
In the same way,
(vtt,φ)+(a4vxx,φxx)−(a3uxx,φxx)+12((u2x+v2x)vx,φx)+ω2(Rvx,φx)−(p2,φ)=0 | (98) |
for arbitrary
Synthesizes the above analysis, there exists
u,v∈L∞(0,T;H20(Ω)),ut,vt∈L∞(0,T;L2(Ω)),utt,vtt∈L∞(0,T;H−2(Ω)). |
(ⅲ) Now let's prove the initial conditions.
Since,
uμ(x,0)→u(x,0) weakly inL2(Ω). | (99) |
Otherwise,
um(x,0)→u0(x) in H20(Ω). | (100) |
Combining identities (99) and (100), we can get
u(x,0)=u0(x). |
Next, according to (83), we can obtain
(uμ,t,φk)→(ut,φk) weakly * in L∞(0,T), |
(uμ,tt,φk)→(utt,φk) weakly * in L∞(0,T). |
Then, we can discover
(uμ,t(x,0),φk)→(ut,φk)|t=0=(ut(x,0),φk). | (101) |
Otherwise,
(um,t(x,0),φk)→(u1(x),φk), | (102) |
Comparing identities (101) and (102), we can get
(ut(x,0),φk)=(u1(x),φk), for arbitrary k. |
So
ut(x,0)=u1(x). |
In the same way, we can obtain
v(x,0)=v0(x),vt(x,0)=v1(x). |
Proof. Differentiating the first equation of (45) with respect to
(uttt,utt)+((a6utxx)xx,utt)−((a3vtxx)xx,utt)−12(((u2x+v2x)utx)x,utt)−((u2xutx)x,utt)−((uxvxvtx)x,utt)−2ωωt((Rux)x,utt)−ω2((Rutx)x,utt)−(p1,t,utt)=0. | (103) |
Differentiating the second equation of (45) with respect to
(vttt,vtt)+((a4vtxx)xx,vtt)−((a3utxx)xx,vtt)−12(((u2x+v2x)vtx)x,vtt)−((v2xvtx)x,vtt)−((uxvxutx)x,vtt)−2ωωt((Rvx)x,vtt)−ω2((Rvtx)x,vtt)−(p2,t,vtt)=0. | (104) |
Summing the equation (103) and (104), we discover after integrating by parts:
12ddt(||utt||2L2(Ω)+||vtt||2L2(Ω))+12ddt(||√a6utxx||2L2(Ω)+||√a4vtxx||2L2(Ω))−ddt||a3utxxvtxx||L1(Ω)+ω22ddt||√R(x)utx||2L2(Ω)+ω22ddt||√R(x)vtx||2L2(Ω)=12((u2x+v2x)utxx,utt)+(u2xutxx,utt)+(uxvxvtxx,utt)+12((u2x+v2x)vtxx,vtt)+(v2xvtxx,vtt)+(uxvxutxx,vtt)+3(uxuxxutx,utt)+(vxvxxutx,utt)+(uxxvxvtx,utt)+(uxvxxvtx,utt)+(uxuxxvtx,vtt)+3(vxvxxvtx,vtt)+(uxxvxutx,vtt)+(uxvxxutx,vtt)+2ωωt((Rxux,utt)+(Ruxx,utt)+(Rxvx,vtt)+(Rvxx,vtt))+(p1,t,utt)+(p2,t,vtt). | (105) |
Obviously, by Young inequality and Hölder inequality, we have
12ddt(||utt||2L2(Ω)+||vtt||2L2(Ω))+12ddt(||√a6utxx||2L2(Ω)+||√a4vtxx||2L2(Ω))−ddt||a3utxxvtxx||L(Ω)+ddt[ω22(||√Rutx||2L2(Ω)+||√Rvtx||2L2(Ω))]≤C(||ut||2H20(Ω)+||vt||2H20(Ω)+||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H20(Ω)+||v||2H20(Ω)+1). | (106) |
Next integrate (106) with respect to
||utt||2L2(Ω)+||vtt||2L2(Ω)+||√a6utxx||2L2(Ω)+||√a4vtxx||2L2(Ω)−2||a3utxxvtxx||L(Ω)+ω2(||√Rutx||2L2(Ω)+||√Rvtx||2L2(Ω))≤C∫t0||ut||2H20(Ω)+||vt||2H20(Ω)+||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H20(Ω)+||v||2H20(Ω)dt+||utt(x,0)||2L2(Ω)+||utx(x,0)||2L2(Ω)+||utxx(x,0)||2L2(Ω)+||vtt(x,0)||2L2(Ω)+||vtx(x,0)||2L2(Ω)+||vtxx(x,0)||2L2(Ω)+C. | (107) |
Since
||utxx(x,0)||L2(Ω),||vtxx(x,0)||L2(Ω),||utx(x,0)||L2(Ω),||vtx(x,0)||L2(Ω), | (108) |
are bounded. On the other hand, multiplying the first equation and second equation of (45) by
||utt(x,0)||L2(Ω)≤C(||u0||H4(Ω)+||v0||H4(Ω))+C≤C, |
||vtt(x,0)||L2(Ω)≤C(||u0||H4(Ω)+||v0||H4(Ω))+C≤C. |
where the condition (52), Young inequality and Poincaré inequality are used.
As before, we have
||utt||2L2(Ω)+||vtt||2L2(Ω)+||ut||2H20(Ω)+||vt||2H20(Ω)≤C∫t0(||utt||2L2(Ω)+||vtt||2L2(Ω)+||ut||2H20(Ω)+||vt||2H20(Ω))dt+C. | (109) |
Applying the Gronwall inequality to (109) gives
ut,vt is bounded in L∞(0,T;H20(Ω)),utt,vtt is bounded in L∞(0,T;L2(Ω)). | (110) |
Moreover, the weak solutions of (45) satisfies
ut∈L∞(0,T;H20(Ω)), |
vt∈L∞(0,T;H20(Ω)), |
utt∈L∞(0,T;L2(Ω)), |
vtt∈L∞(0,T;L2(Ω)). |
Proof. Multiplying the first equation and the second equation of (45) by
(utt,−φ2utxx)+(vtt,−φ2vtxx)+((a6uxx)xx,−φ2utxx)+((a4vxx)xx,−φ2vtxx)−((a3vxx)xx,−φ2utxx)−((a3uxx)xx,−φ2vtxx)−12(((u2x+v2x)ux)x,−φ2utxx)−12(((u2x+v2x)vx)x,−φ2vtxx)−ω2((Rux)x,−φ2utxx)−ω2((Rvx)x,−φ2vtxx)−(p1,−φ2utxx)−(p2,−φ2vtxx)=0. | (111) |
Thanks to integration by parts and the properties of
I1+I2+I3+I4+I5=0. | (112) |
where
I1=(utt,−φ2utxx)+(vtt,−φ2vtxx)I2=((a6uxx)xx,−φ2utxx)+((a4vxx)xx,−φ2vtxx)I3=−((a3vxx)xx,−φ2utxx)−((a3uxx)xx,−φ2vtxx)I4=−12(((u2x+v2x)ux)x,−φ2utxx)−12(((u2x+v2x)vx)x,−φ2vtxx)I5=−(p1,−φ2utxx)−(p2,−φ2vtxx). |
To conclude, we need to estimate each of
I1=12ddt(||φutx||2L2(Ω)+||φvtx||2L2(Ω))+2(φφxutt,utx)+2(φφxvtt,vtx)≥12ddt(||φutx||2L2(Ω)+||φvtx||2L2(Ω))−C | (113) |
For
I2=12ddt(||√a6φuxxx||2L2(Ω)+||√a4φvxxx||2L2(Ω))+(a6,xuxx,φ2utxxx)+(a4,xvxx,φ2vtxxx)+2(a6uxxx,φφxutxx)+2(a4vxxx,φφxvtxx)+2(a6,xuxx,φφxutxx)+2(a4,xvxx,φφxvtxx)≥12ddt(||√a6φuxxx||2L2(Ω)+||√a4φvxxx||2L2(Ω))−(a6,xuxxx,φ2utxx)−(a4,xvxxx,φ2vtxx)+2(a6uxxx,φφxutxx)+2(a4vxxx,φφxvtxx)−C≥12ddt(||√a6φuxxx||2L2(Ω)+||√a4φvxxx||2L2(Ω))−C||φuxxx||2L2(Ω)−C||φvxxx||2L2(Ω)−C | (114) |
where Lemma 3.1 and (52) are used. Similarly from (114), we deduce
I3=−(a3vxxx,φ2utxxx)−(a3uxxx,φ2vtxxx)−(a3,xvxx,φ2utxxx)−(a3,xuxx,φ2vtxxx)−2(a3vxxx,φφxutxx)−2(a3uxxx,φφxvtxx)−2(a3,xvxx,φφxutxx)−2(a3,xuxx,φφxvtxx)=−ddt||a3φ2uxxxvxxx||L1(Ω)−(a3,xvxx,φ2utxxx)−(a3,xuxx,φ2vtxxx)−2(a3vxxx,φφxutxx)−2(a3uxxx,φφxvtxx)−2(a3,xvxx,φφxutxx)−2(a3,xuxx,φφxvtxx)=−ddt||a3φ2uxxxvxxx||L1(Ω)+(a3,xxvxx,φ2utxx)+(a3,xxuxx,φ2vtxx)+(a3,xvxxx,φφxutxx)+(a3,xuxxx,φφxvtxx)+2(a3,xvxx,φφxutxx)+2(a3,xuxx,φφxvtxx)−2(a3vxxx,φφxutxx)−2(a3uxxx,φφxvtxx)−2(a3,xvxx,φφxutxx)−2(a3,xuxx,φφxvtxx)≥−ddt||a3φ2uxxxvxxx||L1(Ω)−C||φuxxx||2L2(Ω)−C||φvxxx||2L2(Ω)−C | (115) |
By using Lemma 3.1 and (52), we find
I4=−C||utxx||2L2(Ω)−C||vtxx||2L2(Ω)−C||uxx||2L2(Ω)−c||vxx||2L2(Ω)≥−C | (116) |
I5=−C(||ut||2H20(Ω)+||vt|2H20(Ω)+||u||2H20(Ω)+||v||2H20(Ω))−C≥−C | (117) |
Putting (113)-(117) into (112), this yields
12ddt(||φutx||2L2(Ω)+||φvtx||2L2(Ω)+||√a6φuxxx||2L2(Ω)+||√a4φvxxx||2L2(Ω) −2||a3φ2uxxxvxxx||L1(Ω))≤C(||φuxxx||2L2(Ω)+||φvxxx||2L2(Ω))+C | (118) |
Integrating with respect to
||φutx||2L2(Ω)+||φvtx||2L2(Ω)+||φuxxx||2L2(Ω)+||φvxxx||2L2(Ω) |
≤C∫t0||φuxxx||2L2(Ω)+||φvxxx||2L2(Ω)dt+C(||u1||2H1(Ω)+||v1||2H1(Ω)+||u0||2H3(Ω)+||v0||2H3(Ω))+C | (119) |
Thanks to the Gronwall inequality, we deduce
||φuxxx||2L2(Ω)+||φvxxx||2L2(Ω)≤C | (120) |
Proof. Utilizing to the first and second equations of (45), we can deduce
((a6a4−a23)uxxxx,uxxxx)=(−a4utt−a3vtt+ω2(R(a4ux+a3vx))x,uxxxx)+12(((u2x+v2x)(a4ux+a3vx))x,uxxxx)+((a4p1+a3p2),uxxxx). | (121) |
Thanks to (50) and Hölder inequality, we see that
||uxxxx||L2(Ω)≤C(||utt+vtt+ut+vt||L2(Ω)+||u+v||H20(Ω))≤C. | (122) |
Furthermore,
||uxxx||L2(Ω)≤C||uxxxx||47L2(Ω)||u||37L2(Ω)+C||u||L2(Ω)≤C, | (123) |
where Gagliardo-Nirenberg inequality for bounded domains is used. Similarly, we deduce
||vxxx||L2(Ω)≤C,||vxxxx||L2(Ω)≤C. | (124) |
Combining with the conclusions in Theorem (4.2) and Theorem (4.3), we can find that
u,v∈L∞(0,T;H4(Ω)∪H20(Ω)). |
Proof. Denote
{˜u,˜v}=π({˜a3,˜a4,˜a6,˜ω,˜θ,˜u0,˜v0,˜u1,˜v1}), |
η=u−˜u, ζ=v−˜v. |
Then
12ddt(||ηt||2L2(Ω)+||ζt||2L2(Ω)+||√a6ηxx||2L2(Ω) +||√a4ζxx||2L2(Ω)−2||a3ηxxζxx||L(Ω))=12(((u2x+v2x)ux−(˜u2x+˜v2x)˜ux)x,ηt)+12(((u2x+v2x)vx−(˜u2x+˜v2x)˜vx)x,ζt)+ω2(((R(x)ηx)x,ηt)+((R(x)ζx)x,ζt))+ω2(η,ηt)+(¯p1,ηt)+(¯p2,ζt) |
−((a6−˜a6)˜uxx,ηtxx)−((a4−˜a4)˜vxx,ζtxx)+((a3−˜a3)˜vxx,ηtxx)+((a3−˜a3)˜uxx,ζtxx)−(ω2−˜ω2)(R˜ux,ηtx)−(ω2−˜ω2)(R˜vx,ζtx)+(ω2−˜ω2)(˜u,ηt)+(ωt−˜ωt)(R0+x,ηt). | (125) |
In equation (125), the nonlinear term satisfies
((u2x+v2x)ux−(˜u2x+˜v2x)˜ux)x=((u2x+v2x)ηx)x+((u2x+v2x−˜u2x−˜v2x)˜ux)x=((u2x+v2x)ηx)x+(((ux+˜ux)ηx+(vx+˜vx)ζx)˜ux)x=2(uxuxx+vxvxx)ηx+(u2x+v2x)ηxx+((ux+˜ux)ηxx+(vx+˜vx)ζxx)˜ux+((uxx+˜uxx)ηx+(vxx+˜vxx)ζx)˜ux+((ux+˜ux)ηx+(vx+˜vx)ζx)˜uxx. | (126) |
By Hölder inequality and Sobolev inequality, we can obtain
(((u2x+v2x)ux−(˜u2x+˜v2x)˜ux)x,ηt)≤C||uxx+vxx||L2(Ω)||ηx||L∞(Ω)||ηt||L2(Ω)+C||ηxx||L2(Ω)||ηt||L2(Ω)+||ηxx+(||uxx+˜uxx||L2(Ω)||ηx||L∞(Ω)+||vxx+˜vxx||L2(Ω)||ζx||L∞(Ω))||ηt||L2(Ω)+(|ηx||L∞(Ω)+||ζx||L∞(Ω))||˜uxx||L2(Ω)||ηt||L2(Ω)+ζxx||L2(Ω)||ηt||L2(Ω)≤C||η||H20(Ω)||ηt||L2(Ω)+C||η||H20(Ω)||ηt||L2(Ω)+C||η||H20(Ω)||ηt||L2(Ω)+C||ζ||H20(Ω)||ηt||L2(Ω)≤C(||ηt||2L2(Ω)+||η||2H20(Ω)+||ζ||2H20(Ω)). | (127) |
In the same way, we have
(((u2x+v2x)vx−(˜u2x+˜v2x)˜vx)x,ζt)≤C(||ζt||2L2(Ω)+||η||2H20(Ω)+||ζ||2H20(Ω)). | (128) |
On the other hand, we easily have
(¯p1,ηt)+(¯p2,ζt)≤C(||θ−˜θ||L1(Ω)+||ηt||2L2(Ω)+||ζt||2L2(Ω)+||η||2H10(Ω)+||ζ||2H10(Ω)). | (129) |
Substituting (127)-(129) into (125), we deduce
||ηt||2L2(Ω)+||ζt||2L2(Ω)+||η||2H20(Ω)+||ζ||2H20(Ω)≤C∫t0||ηt||2L2(Ω)+||ζt||2L2(Ω)+||η||2H20(Ω)+||ζ||2H20(Ω)dt+||u1−˜u1||2L2(Ω)+||v1−˜v1||2L2(Ω)+||u0−˜u0||2H20(Ω)+||v0−˜v0||2H20(Ω)+C(5∑i=3||ai−˜ai||L∞(Ω)+||θ−˜θ||L1(Ω)+||ω−˜ω||W1,1(0,T)). | (130) |
where we used the inequalities
||˜u||L∞(Q)≤C,||˜ux||L∞(Q)≤C,||˜v||L∞(Q)≤C,||˜vx||L∞(Q)≤C,||˜uxx||L∞(0,T,L2(Ω))≤C,||˜vxx||L∞(0,T,L2(Ω))≤C,||ηtx||L∞(Q)≤C,||ζtx||L∞(Q)≤C, |
which are deduced from Theorem 4.2 and Theorem 4.3.
Proof. Assume
{˜φk}∞k=1is an orthogonal basis ofH2f(Ω), | (131) |
and
{˜φk}∞k=1is an orthonormal basis ofL2(Ω). | (132) |
Fix a positive integer
um=m∑k=1˜dk0m(t)˜φk, vm=m∑k=1˜dk1m(t)˜φk, | (133) |
where we intend to select the coefficients
˜dk0m(0)=(u0,˜φk),˜dk1m(0)=(v0,˜φk),k=1,⋯,m, | (134) |
˜dk0m,t(0)=(u1,˜φk),˜dk1m,t(0)=(v1,˜φk),k=1,⋯,m, | (135) |
and
(um,tt,˜φk)+(a6um,xx,˜φk,xx)−(a3vm,xx,˜φk,xx)+12((u2m,x+v2m,x)um,x,˜φk,x)+ω2(Rum,x,˜φk,x)−(p1m,˜φk)−(b1um(l)+e1vm(l))˜φ(l)=0, | (136) |
(vm,tt,˜φk)−(a3um,xx,˜φk,xx)+(a4vm,xx,˜φk,xx)+12((u2m,x+v2m,x)vm,x,˜φk,x)+ω2(Rvm,x,˜φk,x)−(p2m,˜φk)−(b2um(l)+e2vm(l))˜φ(l)=0. | (137) |
As in earlier treatments of C-C boundary condition, we can conclude the following two conclude without difficulty.
(ⅰ) For each integer
(ⅱ)
Then, we proof the following estimate
‖um,t‖2L2(Ω)+‖vm,t‖2L2(Ω)+‖um‖2H2f(Ω)+‖vm‖2H2f(Ω)≤C(‖u1‖2L2(Ω)+‖v1‖2L2(Ω)+‖u0‖2H2f(Ω)+‖v0‖2H2f(Ω))+C | (138) |
Similarly from (75), we can deduce
12ddt{||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||√a6um,xx||2L2(Ω)+||√a4vm,xx||2L2(Ω)+ω22(||√Rum,x||2L2(Ω)+||√Rvm,x||2L2(Ω))}−ddt||a3um,xxvm,xx||L1(Ω)+18ddt||u2m,x+v2m,x||2L2(Ω)−12ddt{b1u2m(l)+e2v2m(l)+2e1um(l)vm(l)}=ωωt(||√Rum,x||2L2(Ω)+||√Rvm,x||2L2(Ω))+(p1m,um,t)+(p2m,vm,t), | (139) |
where
Then we integrate (139) with respect to
||um,t||2L2(Ω)+||vm,t||2L2(Ω)+||um||2H2f(Ω)+||vm||2H2f(Ω)−b1u2m(l)−e2v2m(l)−2e1um(l)vm(l)≤C(||um,t(0)||2L2(Ω)+||vm,t(0)||2L2(Ω)+||um(t)||2H2f(Ω)+||vm(t)||2H2f(Ω))−b1u20m(l)−e2v20m(l)−2e1u0m(l)v0m(l)+C∫t0||um,t||2L2(Ω)+||vm,t||2L2(Ω)dt+||um||2H2f(Ω)+||vm||2H2f(Ω)+C. | (140) |
By simple calculation, we deduce
−b1u2m(l)−e2v2m(l)−2e1um(l)vm(l)≥0. | (141) |
where
On the other hand, according to
u0m(x),v0m(x)∈C(¯Ω). |
Thus,
u0m(l),v0m(l)≤C. | (142) |
Substituting (141) and (142) into (140), applying Gronwall inequality, we can deduce (138).
Now we pass to limits in our Galerkin approximations, applying estimate (138), we can discover (92), (93), (95). In order to complete the proof of the theorem, we just have to proof
uμ(l)→u(l),vμ(l)→v(l), strongly in L∞(0,T). | (143) |
where
To verify this, recalling (138), we observe that
uμ→u,vμ→v strongly in C(0,T;H1f(Ω)) | (144) |
where the Corollary 4 of Chapter 8 in [30] is used.
Furthermore, thanks to the conditions
||uμ(l)−u(l)||L∞(0,T)=||(uμ(l)−u(l))−(uμ(0)−u(0))||L∞(0,T)=||∫l0(uμ(x)−u(x))xdx||L∞(0,T)≤√l||(uμ(x)−u(x))x||L∞(0,T;L2(Ω))≤√l||uμ(x)−u(x)||L∞(0,T;H1f(Ω)) | (145) |
Thanks to (144), we can deduce
uμ(l)→u(l) strongly in L∞(0,T), |
Similarly, we have
vμ(l)→v(l) strongly in L∞(0,T). |
Proof. Similarly as (105), we can get
12ddtE+12P1+P2=P3. | (146) |
where
E=||utt||2L2(Ω)+||vtt||2L2(Ω)+||√a6utxx||2L2(Ω)+||√a4vtxx||2L2(Ω)−2||a3utxxvtxx||L(Ω)+ω2||√R(x)utx||2L2(Ω)+ω2||√R(x)vtx||2L2(Ω) |
P1=(((u2x+v2x)ux)t,uttx)+(((u2x+v2x)vx)t,vttx)P2=−(b1u(l)+e1v(l))tutt(l)−(b2u(l)+e2v(l))tvtt(l)P3=2ωωt(((Rux)x,utt)+((Rvx)x,vtt))+||√Rutx||2L2(Ω)+||√Rvtx||2L2(Ω))+(p1,t,utt)+(p2,t,vtt) |
By calculation, we can deduce
P1=12ddt{||√u2x+v2xutx||2L2(Ω)+||√u2x+v2xvtx||2L2(Ω)+2||uxutx||2L2(Ω) +2||vxvtx||2L2(Ω)+4||uxvxutxvtx||L1(Ω)} −3∫l0uxu3tx+vxv3tx+uxutxv2tx+vxvtxu2txdx | (147) |
P2=−12ddt{b1u2t(l)+e2v2t(l)+2e1ut(l)vt(l)} | (148) |
P3≤||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H2f(Ω)+||v||2H2f(Ω) +||utx||2L2(Ω)+||vtx||2L2(Ω) | (149) |
Substituting (147), (148) and (149) into (146), we get
12ddt{E+||√u2x+v2xutx||2L2(Ω)+||√u2x+v2xvtx||2L2(Ω)+2||uxutx||2L2(Ω) +2||vxvtx||2L2(Ω)+4||uxvxutxvtx||L1(Ω) −(b1u2t(l)+e2v2t(l)+2e1ut(l)vt(l))}≤||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H2f(Ω)+||v||2H2f(Ω)+||utx||2L2(Ω)+||vtx||2L2(Ω) +3∫l0uxu3tx+vxv3tx+uxutxv2tx+vxvtxu2txdx | (150) |
Then integrate (150) with respect to
E≤∫t0||utt||2L2(Ω)+||vtt||2L2(Ω)+||u||2H2f(Ω)+||v||2H2f(Ω) +||utx||2L2(Ω)+||vtx||2L2(Ω)dt +3∫t0∫l0uxu3tx+vxv3tx+uxutxv2tx+vxvtxu2txdxdt +||utt(x,0)||2L2(Ω)+||u1||2H2f(Ω)+||vtt(x,0)||2L2(Ω)+||v1||2H2f(Ω). |
where
−b1u2t(l)−e2v2t(l)−2e1ut(l)vt(l)≥0 |
and
||uxutx||2L2(Ω)+||vxvtx||2L2(Ω)+2||uxvxutxvtx||L1(Ω)≥0. |
are used.
Similarly as (108), we discover
||utt(x,0)||L2(Ω),||vtt(x,0)||L2(Ω)≤c. | (151) |
According to (151), we get
E≤c∫t0||utt||2L2(Ω)+||vtt||2L2(Ω)+||utxx||2L2(Ω)+||vtxx||2L2(Ω)dt+3∫t0∫l0uxu3tx+vxv3tx+uxutxv2tx+vxvtxu2txdxdt+c. |
By Hölder inequality and Sobolev inequality, we have
E≤c∫t0{||utt||2L2(Ω)+||utxx||2L2(Ω)+||utxx||3L2(Ω)+||utxx||4L2(Ω)||vtt||2L2(Ω)+||vtxx||2L2(Ω)+||vtxx||3L2(Ω)+||vtxx||4L2(Ω)}dt+c. | (152) |
Then, we obtain Theorem 5.3 by Gronwall inequality.
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